Answer
$\dfrac{\pi^2}{12} $
Work Step by Step
We have $1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......=\dfrac{\pi^2}{6} ~~~~~(a)$
Let us consider that $S=1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+......~~~~(b)$
Next, we will subtract equation (b) from equation (a) to obtain:
$\dfrac{\pi^2}{6}-S=(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......)-(1-\dfrac{1}{2^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+......)$
or, $=2 [\dfrac{1}{2^2}+\dfrac{1}{4^2}+....]$
or, $=2 \Sigma_{k=1}^{\infty} \dfrac{1}{(2k)^2}$
or, $=\dfrac{1}{2} \times \dfrac{\pi^2}{6} $
Thus, $S=\dfrac{\pi^2}{6} -\dfrac{\pi^2}{12} =\dfrac{\pi^2}{12} $
